hair-tearing perl question
Dec. 20th, 2006 05:49 pmWhat I want: a subroutine footle such that, if you call footle(a,b) twice with the same a,b, it does nothing the second time
What I did:
But this doesn't work because parameters are passed by value.
But if I call as footle("bootle","bumtrinket",\%isdone), which passes isdone by reference, it still does the footling twice.
Even if I put $_[2]=%done before the end of the subroutine, it still does the footling twice.
And if I put print join "*",(keys %done); at the start of the subroutine, it saysHASH(0x8188110)footling bootle bumtrinket
So how do I really pass the parameter by reference, as if I'd said void footle(int a, int b, set<string>& done) in C++?
What I did:
use strict; sub footle { my ($a,$b,%done) = @_; my $concat = $a.$b; if ($done{$concat} == 0) { print "footling $a $b"; $done{$concat} = 1; } } my %isdone = (); footle("bootle","bumtrinket",%isdone); footle("bootle","bumtrinket",%isdone);
But this doesn't work because parameters are passed by value.
But if I call as footle("bootle","bumtrinket",\%isdone), which passes isdone by reference, it still does the footling twice.
Even if I put $_[2]=%done before the end of the subroutine, it still does the footling twice.
And if I put print join "*",(keys %done); at the start of the subroutine, it says
So how do I really pass the parameter by reference, as if I'd said void footle(int a, int b, set<string>& done) in C++?